Power (Mechanical)

Claimed by Aeze3

This topic covers Power.

The Main Idea

Power is the rate of doing work or the amount of energy consumed over an interval of time.

A Mathematical Model

When a force is applied over a distance in a unit of time, power is calculated by

[math]\displaystyle{ power = \frac{F \Delta r}{\Delta t} = \frac{W}{\Delta t} }[/math]

where F is force, Δr is displacement, Δt is the duration of time and W is work.

It then follows that instantaneous power is

[math]\displaystyle{ power = F\cdot v }[/math]

where v is velocity.

The SI unit for power is watts (J/s)

A Computational Model

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Examples

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Simple

A certain motor is capable of doing 3000 J of work in 12 s What is the power output of this motor?

[math]\displaystyle{ power = \frac{W}{\Delta t} = \frac{3000 J}{12 s} = 250 watts }[/math]

Middling

Here are questions dealing with human power. (a) If you follow a diet of 2000 food calories per day (2000 kC), what is your average rate of energy consumption in watts (power input)? (A food or “large” calorie is a unit of energy equal to 4.2 J; a regular or “small” calorie is equal to 4200 J.) (b) How many days of a diet of 2000 large calories are equivalent to the gravitational energy change from sea level to the top of Mount Everest, 8848 m above sea level? Assume your weight is 58 kg. (The body is not anywhere near 100% efficient in converting chemical energy into change in altitude. Also note that this is in addition to your basal metabolism.)

(a) [math]\displaystyle{ power = \frac{W}{\Delta t} = \frac{2000 kC}{day} \cdot \frac{4200 J}{1 kC} \cdot \frac{1 day}{24 h} \cdot \frac{1 h}{3600 s} = 97.2 watts }[/math]

(b) [math]\displaystyle{ {\frac{97.2 J}{s}} \cdot \frac{3600 s}{1 h} \cdot \frac{24 h}{1 day} = 8398080 J/day }[/math]

[math]\displaystyle{ \Delta U_g = mg\Delta y = (58 kg)(9.8 m/s^2)(8848 m - 0 m) = 5029203.2 J }[/math]

[math]\displaystyle{ \Delta t = \frac{W}{power} = \frac{5029203.2 J}{8398080 J/day} = 0.599 days }[/math]

Difficult

A bicyclist is going up an inclined slope with an angle [math]\displaystyle{ \alpha }[/math] = 2.9 degrees by a uniform speed of 27 km/h. The magnitude of the air resistance force is given by [math]\displaystyle{ F_{air} = kv^2 }[/math] kgs/m where the numerical value of [math]\displaystyle{ k = 0.3 }[/math] if the unit of the speed is m/s and the unit of the resistance force is the newton (N). The mass of the bicyclist including the bike is 70 kg. Do not consider the rolling resistance. (a) What forward force exerted on the bike by the road is needed to make the bicyclist move with constant speed? (b) How much work does the bicyclist do when riding a distance of 1200 m? (c) What is the power of the bicyclist during the ride? Assume there is no loss of mechanical energy.

[math]\displaystyle{ F_g }[/math] = weight

N = normal force exerted on the bike by the road

[math]\displaystyle{ F_{odp} }[/math] = air resistance

F = unknown forward force exerted on the bike by the road

(a) [math]\displaystyle{ x-components: -F_gsin\alpha -F_{odp} + F = 0 }[/math]

[math]\displaystyle{ y-components: N -F_gcos\alpha = 0 }[/math]

[math]\displaystyle{ F = F_gsin\alpha + F_{odp} = mgsin\alpha + kv^2 }[/math]

[math]\displaystyle{ v = \frac{27 km}{h} \cdot \frac{1000 m}{1 km} \cdot \frac{1 h}{3600 s} = 7.5 m/s }[/math]

[math]\displaystyle{ F = (70 kg)(9.8 m/s^2)sin2.9 + (0.3 kgs/m)(7.5 m/s)^2 = 51.6 N }[/math]

(b) [math]\displaystyle{ W = F\Delta r = (51.6 N)(1200 m) = 61920 J }[/math]

(c) [math]\displaystyle{ power = F\cdot v = (51.6 N)(7.5 m/s) = 387 watts }[/math]

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See also

• Work
• Gravitational Potential Energy
• Newton's Second Law of Motion
• System & Surroundings

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Further reading

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External links

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References

Chabay, Ruth W.; Sherwood, Bruce A. Matter and Interactions, 4th Edition: 1-2. Wiley.

"General Mechanics/Work and Power." - Wikibooks, Open Books for an Open World. Web. General Mechanics/Work and Power

<https://physicstasks.eu/280/bicyclist-going-uphill>.